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Unformatted text preview: University of California San Diego ECE 259A: Solutions to Problem Set #3
1. The order of must divide 2 9 1 511. Hence it is either 7, 73, or 511. Clearly o 7, 7 since otherwise is a root of x 1 which contradicts the fact that its minimal polynomial has degree 9. On the other hand 9 1, so that 72 1 8 8 1, and thus 73 9 1. Hence o 73. q 1 2 q 1 2 q 1 2 1 1 1 Hence the order of is a divisor of q 1 2, and thus cannot be primitive in GF q . (d) Take for instance in GF 2 m . Then 2 is a conjugate of and is therefore primitive. Another option is to consider a field GF 2 m such that 2m 1 is prime (for instance 23 , 25 , or 27 ) in which every element, except 0 and 1, is primitive. 4. The polynomial f x x 2 x 1 is the only irreducible polynomial of degree 2 over GF 2 . Using this polynomial to construct GF 4 0, 1, , 2 as a set of 2tuples over GF 2 , we obtain the following table: 0 0 00 0 1 10 1 01 2 1 11 One choice of a polynomial of degree 2 which is irreducible over GF 4 is p x x2 x There are several other choices. To see that p x it is indeed irreducible, note that p 0 3. then and must have the same minimal polynomial. In other words, must (a) If p 2 r 1 be a conjugate of which implies that belongs to , p , p , . . . , p , where r is the r m smallest integer such that p (such an integer necessarily exists since p for1 all r p in the field). The only element in the above set which satisfies is p . (b) Clearly 1 and 1 are the square roots of 1 in any field. In a field of characteristic two, we have 1 1. Furthermore, by part (a), square roots are unique in such a field and hence 1 is the only square root of 1. Now let be a primitive element in GF p m , where the m characteristic p is odd. Then p 1 2 is welldefined and is the second square root of 1. m This implies, in particular, that in a field of odd characteristic 1 p 1 2 . (c) Let q pm be the order of the field, where p is odd. From part (b), we have It follows that 13 12 1 1, and therefore o 12 4 4 2 3 2 2 6 1 2 2 4 1 3 3 2 2 2 2 2 1 1 2 13. 2 2 2 1 1 2 2. (a) If f x x3 x2 2 is not irreducible, it must have at least one linear factor. But, since f 0 , f 1 , and f 2 are all nonzero, none of x, x 1, and x 2 is a factor of f x . Hence f x is irreducible over GF 3 . Note, however, that this approach does not apply to polynomials of degree greater than three. (b) 2 1 2 2 2 3 2 2 3 2 2 1. (c) The possible multiplicative orders are all the divisors of 26, namely 1, 2, 13, and 26. (d) Since clearly 2 1, the order of must be either 13 or 26. Since 3 2 2 0 in the field, we deduce that 3 2 2 1 and 4 2 3 2 2 2 1 2 2. Also, since 2 3 2 1, we conclude that 2 1 . Now . , p 1 , p 2 , and p 2 2 . Hence p x has no roots in GF 4 , and thus no linear factors. Using this polynomial to construct GF 16 as a set of 2tuples over GF 4 , we obtain: 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 0 0 1 0 0 1 1 2 1 1 0 0 2 2 1 2 0 0 2 1 2 1 2 2 2 1 2 2 2 2 2 1 1 2 2 1 5 1 1 1 4 1 1 4 1 1 1 in the field at hand. Hence o 1 5 and o 1 5, which implies o 1 15. (c) Since 1 is a primitive element of GF 16 its minimal polynomial g x has to be an irreducible primitive polynomial of degree 4. Furthermore, we have g 1 0 f . Thus g 1 4 3 2 1 0. Let 1, so that 1. Then, g 4 1 4 1 3 1 1 2 1 1 1 1 2 2 1 1 1 4 3 1 Hence g x x4 x3 1. This is an example of a general method which can be used to obtain a primitive irreducible polynomial from a nonprimitive irreducible polynomial. (d) A typical element of GF 2 x f x is given by a 3 3 a2 2 a1 a0 while a typical element of GF 2 x g x is given by b 3 3 b2 2 b1 b0 , where ai , bi GF 2 and 1. Using the latter relation, we find that b 0 , b1 , b2 , b3 GF 2 x g x may be expressed as b3 1 3 Thus the isomorphism : GF 2 x g x GF 2 x f x maps a into b3 b2 b1 b0 , b3 b1 , b3 b2 , b3 GF 2 x f x . b 0 , b1 , b2 , b3 b 3 3 b3 b2 2 b3 b1 b3 b2 b1 b3 3 2 b2 1 2 b1 2 1 b0 1 b2 1 b1 1 b0 b0 &% 0 # ) 5 4 4 3 2 3 2 5. (a) We have x5 1 x 1 f x . Hence if is a root of f x it is also a root of x 5 1, and therefore o 5. Thus cannot be primitive in GF 16 . (b) The order of any element in GF 16 , except 0 and 1, is either 3, 5 or 15. It is clear that 1 3 1 0, since the minimal polynomial of is of degree 4. Furthermore, " ) ) # &% $ &% $ 2 &% # 1 #' where is the root of p x , and 15 2 2 2 2 3 (% ' # &% $ # ! 1. 0 (b) Over GF 3 we have x 6 divisors is 42 16. 1 x2 1 3 x 1 3 x 1 3 , and the number of monic All these elements are roots of M x , and hence deg M x A x l. Now denote i 0 and deg A x l. Since M x is the unique polynomial of minimal Then clearly A degree with coefficients in GF p having as a root, it would suffice to show that the coefficients A x . The coefficients a 0 , a1 , . . . , al of of A x are in GF p in order to prove that M x A x are elementary symmetric functions of the elements of C , given by a0 1 l p p 1
l 1
2 p
3 l 1 a1 p p p
2 2 p l 1 p p 2 3 p
3 l 1 p p
l 1 2 3 al 1 p p 2 p l 1 al 1 We need to show that ai GF p , or equivalently ai ai , for all i 0, 1, . . . , l. We may as well consider ai ai 1 l i , since clearly ai ai and ai GF p iff ai GF p . In general, ai is the sum of all the possible products of l i elements from the set C . Now, since p x y p x p y p in GF pm , we observe that ai is still a sum of the same number of lterms, , and these terms are products of l i elements from C raised to the pth power. Since p the pth power of a product of l i elements of C is still a product of l i elements of C . Furthermore, the set of all such products is clearly invariant under the operation of raising to the pth power (this is equivalent to adding 1 modulo l to the set of integers 0, 1, . . . , l 1 ). It p follows that ai ai and ai , ai GF p for all i. ) H ) H P I H G ) ) H H H H F 2 H al p p p l 1 p p 2 p p p p p p l 2 l 1 p $0E EE $0E EE $0E EE 0$E EE . . . EE G $0E $0E $0E EE EE D E $E0E $0E EE B CA j j 0 x j D l 1 x p j k 0 ak xk l 8 C 9 @ 7 8 5 ) j : j i , p , p , . . . , p 2 l 1 p 7. Let l i . Then l is the smallest integer, such that i are given by l , and the cyclotomic conjugates of p l 2 ) 7 F 3 654 3 6. (a) In GF 2 we have x 6 1 x3 1 2 x 1 2 x2 x 1 2 . Thus the factorization of x6 1 contains two irreducible factors, each of multiplicity 2. It follows that any monic divisor of x6 1 over GF 2 has the form x 1 i x2 x 1 j , where i, j 0, 1, 2 . 2 6 This gives 3 9 monic divisors, including the trivial ones 1 and x 1. ...
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 Fall '08
 Prof.AlexanderVardy

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